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2 years ago

Solve for y 3x + 1/5y = 7

Mathematics

1 answer:

jarptica [38.1K]2 years ago

4 0

I think it is y=35 because u multiply both sides by 5 move the variable right then the final solution is

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PYTHAGOREAN THEOREM IN 3D URGENT PLEASE PHOTO INSERTED

Andrews [41]

To find the length of the diagonal, we need to calculate the length of the base first and then use the value to calculate the diagonal since we are given the value of height of the triangle.

<h3>Pythagorean Theorem</h3>

This theorem is used to find the missing side of a right angle triangle given that we have the length of two sides.

To calculate the base of the triangle, let's use Pythagorean theorem here

$x^2 = y^2 + z^2\\x^2 = 2^2 + 3^2\\x^2 = 4 + 9 \\x^2 = 13\\x = \sqrt{13} \\x = 3.61$

Having the length of the base, let's consider the height of the triangle which is 6 and substitute the values.

$d^2 = 6^2 + 3.61^2\\d^2 = 36 + 13.0321\\d^2 = 49.0321\\d = \sqrt{49.0321} \\d = 7.00$

From the calculations above, the values of the triangle are

diagonal = 7
base = 3.61
height = 6

Learn more on pythagorean theorem here;

brainly.com/question/231802

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7 0

1 year ago

You can do both problems if want

Agata [3.3K]

A= π r2 and v=3.14d

---Volume(v=3.14d):

3.14(15) equals 47.1

47.1 rounds to 47

nearest hundreth would be 0 that rounds to .1 or .10,which rounds to 47
v=47

---Area(π r2)

3.14(7.5)^2 is 3.14(56.25)
176.625

176.625 rounded to the nearest hundreth is 176.630.

Answer:

The area of a circle with the diameter of 15 is 176.630.

4 0

3 years ago

If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?

oksano4ka [1.4K]

$\boxed{ \ x \geq 1 \ }$ or can be written as $\boxed{ \ [1, \infty) \ }$

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given $\boxed{ \ a(x) = 3x + 1 \ }$ and $\boxed{ \ b(x) = \sqrt{x - 4} \ }$ .

We will form (b o a)(x) and then determine the domain.

$\boxed{ \ (b \circ a)(x) = b(a(x)) \ }$

Replace each appearance of x in b(x) with $\boxed{ \ a(x) = 3x + 1 \ }$ .

$\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }$

Thus, $\boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }$

To be defined, the value under the radical sign must not be negative. Therefore, the domain of $(b \circ a)(x) = \sqrt{3x - 3}$ are processed as follows.

$\boxed{ \ 3x - 3 \geq 0 \ }$

Both sides added by 3.

$\boxed{ \ 3x \geq 3 \ }$

Both sides divided by 3.

$\boxed{ \ x\geq 1 \ }$

Thus, the domain of $(b \circ a)(x) = \sqrt{3x - 3}$ is $\boxed{ \ x \geq 1 \ }$ or can be written as $\boxed{ \ [1, \infty) \ }$

<h3>Learn more</h3>

If f(x) = x² – 2x and g(x) = 6x + 4, for which value of x does (f o g)(x) = 0? brainly.com/question/1774827
Solve for the value of the function composition brainly.com/question/2142762
Look for rotation rules in the transformation brainly.com/question/2992432

Keywords: composition of function, if a(x) = 3x + 1, and, b(x) = √(x-4), what is the domain of, (b o a)(x), b(a(x)), defined, the value, under the radical sign, must not be negative,